when some NaCl was dissolved in water, the freezing point depression was nmerically equal to twice the molal f.p. depression constant. The relative lowering of vapour pressure of the solution in nearly :

Options

A. 0.036
B. 0.018
C. 0.0585
D. 0.072

Answer

To solve the problem, we need to find the relative lowering of vapor pressure when NaCl is dissolved in water, given that the freezing point depression is numerically equal to twice the molal freezing point depression constant.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The freezing point depression (ΔTf) is given as numerically equal to twice the molal freezing point depression constant (Kf).
- Therefore, we can write:
\[ \Delta T_f = 2 K_f \]

2. **Use the Freezing Point Depression Formula**:
- The formula for freezing point depression is:
\[ \Delta T_f = K_f \times m \]
- Where \( m \) is the molality of the solution. From the first step, we can equate:
\[ 2 K_f = K_f \times m \]

3. **Solve for Molality (m)**:
- Dividing both sides by \( K_f \) (assuming \( K_f \neq 0 \)):
\[ 2 = m \]
- Thus, the molality \( m \) is 2 mol/kg.

4. **Calculate the Number of Moles of Solute (NaCl)**:
- The molality \( m \) is defined as:
\[ m = \frac{n_1}{W_2} \times 1000 \]
- Where \( n_1 \) is the number of moles of solute and \( W_2 \) is the weight of the solvent in kg. Rearranging gives:
\[ n_1 = m \times W_2 / 1000 \]
- Substituting \( m = 2 \):
\[ n_1 = 2 \times W_2 / 1000 \]

5. **Calculate the Number of Moles of Solvent (Water)**:
- The molecular weight of water (solvent) is 18 g/mol. Thus, the number of moles of solvent \( n_2 \) is:
\[ n_2 = \frac{W_2}{18} \]

6. **Calculate the Relative Lowering of Vapor Pressure**:
- The relative lowering of vapor pressure is given by:
\[ \text{Relative lowering} = \frac{n_1}{n_1 + n_2} \]
- Substituting \( n_1 \) and \( n_2 \):
\[ \text{Relative lowering} = \frac{2 \times W_2 / 1000}{2 \times W_2 / 1000 + W_2 / 18} \]

7. **Simplify the Expression**:
- Factor out \( W_2 \):
\[ \text{Relative lowering} = \frac{2/1000}{2/1000 + 1/18} \]
- To combine the terms, find a common denominator (1000 and 18):
\[ = \frac{2/1000}{\frac{2 \times 18 + 1000}{1000 \times 18}} = \frac{2 \times 18}{2 \times 18 + 1000} \]

8. **Calculate the Final Value**:
- Calculate \( 2 \times 18 = 36 \):
\[ \text{Relative lowering} = \frac{36}{36 + 1000} = \frac{36}{1036} \approx 0.0347 \approx 0.036 \]

### Final Answer:
The relative lowering of vapor pressure of the solution is approximately **0.036**.